A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3*x^k]*(1-x)^(3n+1), read by rows of k=0..2n terms.

1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 54, 405, 760, 405, 54, 1, 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 490, 35623, 818300, 7917371, 37215794, 91789005, 123519792, 91789005, 37215794, 7917371, 818300, 35623, 490, 1, 1, 704, 73200, 2430400, 34657700, 246781248, 955910032, 2116980800, 2751843600, 2116980800, 955910032, 246781248, 34657700, 2430400, 73200, 704, 1

Offset: 0

Author

Paul D. Hanna, Oct 30 2010

Keywords

• nonn
• tabf

Examples

			Triangle begins:
 1;
 1, 4, 1;
 1, 20, 48, 20, 1;
 1, 54, 405, 760, 405, 54, 1;
 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1; ...
Row g.f.s begin:
 (1) = (1-x)*(1 + x + x^2 + x^3 + x^4 +...);
 (1 + 4*x + x^2) = (1-x)^4*(1 + 2^3*x + 3^3*x^2 + 4^3*x^3 +...);
 (1 + 20*x + 48*x^2 + 20*x^3 + x^4) = (1-x)^7*(1 + 3^3*x + 6^3*x^2 +...);
 (1 + 54*x + 405*x^2 + 760*x^3 + 405*x^4 + 54*x^5 + x^6) = (1-x)^10*(1 + 4^3*x + 10^3*x^2 + 20^3*x^3 + 35^3*x^4 +...); ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..1088, as a flattened triangle of rows 0..32
• Ilia Gaiur, Vladimir Rubtsov, and Duco van Straten, Product formulas for the Higher Bessel functions, arXiv:2405.03015 [math.AG], 2024. See p. 18.
• Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 14.

Crossrefs

Cf. A181543, A181545, A006480, A002897, A000172.

Cf. A183204 (central terms), A183205.

Programs

Mathematica

t[n_, k_] := SeriesCoefficient[Sum[Binomial[n+j, j]^3*x^j, {j, 0, n+k}]*(1-x)^(3*n+1), {x,0, k}]; Table[t[n, k], {n, 0, 9}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Feb 04 2014, after PARI *)

PARI

{T(n,k)=polcoeff(sum(j=0,n+k,binomial(n+j,j)^3*x^j)*(1-x)^(3*n+1),k)}
for(n=0,10,for(k=0,2*n,print1(T(n,k),", "));print(""))

Formula

Row sums equal A006480(n) = (3n)!/(n!)^3, which is de Bruijn's s(3,n).

From Yahia Kahloune, Jan 30 2014: (Start)

Using these coefficients we can obtain formulas for the sums

Sum_{i=1..n} C(e-1+i,e)^3. Let us define b(k,e,3) = sum_{i=0..k-e} (-1)^i*C(3*e+1,i)*C(k-i,e)^3, where k=e+i.

For example:

b(e,e,3) = 1;

b(e+1,e,3) = (e+1)^3-(3*e+1) = e^2*(e+3);

b(e+2,e,3) = C(e+2,2)^3 - (3*e+1)*(e+1)^3 + C(3*e+1,2);

b(e+3,e,3) = C(e+3,e)^3 - (3*e+1)*C(e+2,e)^3 + C(3*e+1,2)*C(e+1,e)^3 - C(3*e+1,3);

b(e+4,e,3) = C(e+4,e)^3 - (3*e+1)*C(e+3,e)^3 + C(3*e+1,2)*C(e+2,e) - C(3*e+1,3)*C(e+1,e)^3 + C(3*e+1,4).

Then we have the formula: Sum_{i=1..n} C(e-1+i,e)^3 = Sum_{i=0..2*e} b(e+i,e,3)*C(n+e+i,3*e+1).

Example: Sum_{i=1..7} C(2+i,3)^3 = C(10,10) + 54*C(11,10) + 405*C(12,10) + 760*C(13,10) + 405*C(14,10) + 54*C(15,10) + C(16,10) = 820260. (End)

Let E be the operator D*x*D*x*D, where D denotes the derivative operator d/dx. Then (1/(n)!^3) * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(3*n+1) = Sum_{k >= 0} binomial(n+k, k)^3*x^k. For example, when n = 2 we have (1/2!)^3*E^3(1/(1 - x)) = (1 + 20 x + 48 x^2 + 20 x^3 + x^4)/(1 - x)^7. - Sergii Voloshyn, Dec 03 2024